maths p32 answers

[redapple20]

Can u tel me the position vector of p?

 

[OrlandoBloom<3]

Please send me the email as well My paper was extremely bad :/ But I am pretty sure if we had enough time everyone would be able to do well . Because for me eevrytime I wasnt getting the answer right I wouldnt get time to correct it , just move on to the next because of the fear of time ending :/ I felt like I was so lucky giving o levels last year because all the papers were so easy and now I feel like we are the unluckiest giving A levels this year because almost all the papers I have given till now are prooving to be so harddd :/ Anyways did you find the differential equation weird ? I was unable to solve it because I have never come across a question like that ! Was it a repeated question from the past years ? Because I am sure I solved all papers and didnt come across an equation like that :/

 

[Zika999]

If I lost 15 marks, would I still get an A? Please reply ASAP ​

 

[sarahkhan]

Yes, if the threshold is down to 60 for an A

 

[gudboy01]

Are you sure of the lambda thing? I think it was 3? And i'm not sure about the other value of lambda?

Does anyone remember the last question? NOT THE ANSWER, THE QUESTION?

ohh..yah i too got the value of lamda 3...

 

[aaditya menon]

Was there a need to find the terms in x^2 (like the -1/8x^2 and 3/8x^2) because (1-0.5x)(1-0.5x) will give you the terms in x and x^2.

Yes, coz they too give you an answer i x^2 (when multiplied by 1)

 

[sweetyrox]

Can anybody send me a copy of the p32 also plzzz
My email add is: [email protected]

 

[Ahmadbasharat]

In the 1st question, the value of x was 2.30 and in the vectors ques ii) part, the lamda was 2. correct me if i am wrong.

can you plz tell me what was the second part of vectores qs

 

[earthling]

uhm weren't there 3 stationary points in q 6, as in pie/2, pie/12 and 5pie/12?

But doesn't lne get cancelled?

I got y= x -ln(3/2)

 

[ffaadyy]

But doesn't lne get cancelled?

I got y= x -ln(3/2)

The question was dy/dx=e^(2x+y) right?

dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx

Integrate both the sides.

-e^(-y) = [e^(2x)/2] + c

Put x=0 and y=0 to find the value of 'c'.

-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c

Put back this value of 'c' in the integrated solution.

-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)

Multiply the equation by a '-' sign.

- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]

Put 'ln' on both the sides.

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}

Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.

 

[jumana94]

send me the paper too please, [email protected] ..
id really appreciate it

 

[Saiyan]

ohh..yah i too got the value of lamda 3...

No it was 2

 

[Ahmed Tariq]

Your binomial expansion is wrong. I do not remember the answer though

His binomial expansion is correct.

 

[Ahmed Tariq]

Hey guys, here are some of the answers my tutor got. Please remember that even a math tutor could make a mistake, so the answers below may not be 100% correct. Good luck to all of us!

1. x=2.3
2.ii. The root is 1.32
3. 1 - x + (1/2)x^2
4. pi/2, 7pi/6 ---Note: not 100% sure about this one because we did it really fast, since I didn't solve it completely
5. expression for differential is -y=ln[(-1/2)e^2x+(3/2)] -> y= -ln[-1/2 e^2x + 3/2]
6. i. angles are pi/2, pi/12 and 5pi/12 (3 angles)
ii. the point is a maximum point
7. u = -1/2+1/2i
8. ---
9. i.equation of tangent is y= x-1
ii. pi[(1/4)-(e^2/4)]
10. ii. position vector for point of intersection is 5,3,3
we did not solve the last part with the distance for the vector because I didn't attempt the question.

The angles in the trigonometry question had to be given in degrees and not radians. Correct me if I am wrong.

 

[Ahmed Tariq]

Sent.

Could you also kindly send it to [email protected]? Would be appreciated.

 

[Saiyan]

His binomial expansion is correct.

Sorry I forgot! It is right.

 

[Goodfella]

But doesn't lne get cancelled?

I got y= x -ln(3/2)

You're right i got the same answer!

 

[Goodfella]

The question was dy/dx=e^(2x+y) right?

dy/dx=e(2x+y)
dy/dx=e^(2x) x e^( y )
[1/e^( y )] dy = e^(2x) dx
e^(-y) dy = e^(2x) dx

Integrate both the sides.

-e^(-y) = [e^(2x)/2] + c

Put x=0 and y=0 to find the value of 'c'.

-e^(-y) = [e^(2x)/2] + c
-e^(-0) = [e^(0)/2] + c
-1 = (1/2) + c
-1 - (1/2) = c
-3/2 = c

Put back this value of 'c' in the integrated solution.

-e^(-y) = [e^(2x)/2] + c
-e^(-y) = [e^(2x)/2] - (3/2)

Multiply the equation by a '-' sign.

- {-e^(-y) = [e^(2x)/2] - (3/2)}
e^(-y) = (3/2) - [e^(2x)/2]

Put 'ln' on both the sides.

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - [e^(2x)/2]}
y = - ln {(3/2) - [e^(2x)/2]}

Therefore, the answer was 'y = - ln {(3/2) - [e^(2x)/2]}'.

wouldn't e^2x/2 get cancelled when we take ln?

 

[earthling]

wouldn't e^2x/2 get cancelled when we take ln?

Exactly!!!! Ln always cancels an "e"!!!!

 

[ffaadyy]

wouldn't e^2x/2 get cancelled when we take ln?

Exactly!!!! Ln always cancels an "e"!!!!

No, that's wrong. Though I know what you're trying to say. You would've written it like this:

ln e^(-y) = ln {(3/2) - [e^(2x)/2]}
-y = ln {(3/2) - ln [e^(2x)/2]}
-y = ln {(3/2) - (2x)/2}

But this is the wrong method, we can not multiply 'ln' to the numbers inside the bracket. Take this for example, ln (5-4) gives you '0' but if you calculate ln 5 - ln 4, then the answer is different. Therefore, we couldn't have done it your way.